## Tag `0AVF`

Chapter 15: More on Algebra > Section 15.52: Injective modules

Lemma 15.52.4. Let $R$ be a ring. Let $J$ be an $R$-module. The following are equivalent

- $J$ is injective,
- $\mathop{\rm Ext}\nolimits^1_R(R/I, J) = 0$ for every ideal $I \subset R$, and
- for an ideal $I \subset R$ and module map $I \to J$ there exists an extension $R \to J$.

Proof.We have seen the implication (1) $\Leftrightarrow$ (2) in Lemma 15.52.3. In this proof we will show that (1) $\Leftrightarrow$ (3) which is known as Baer's criterion.Assume (1). Given a module map $I \to J$ as in (3) we find the extension $R \to J$ because the map $\mathop{\rm Hom}\nolimits_R(R, J) \to \mathop{\rm Hom}\nolimits_R(I, J)$ is surjective by definition.

Assume (3). Let $M \subset N$ be an inclusion of $R$-modules. Let $\varphi : M \to J$ be a homomorphism. We will show that $\varphi$ extends to $N$ which finishes the proof of the lemma. Consider the set of homomorphisms $\varphi' : M' \to J$ with $M \subset M' \subset N$ and $\varphi'|_M = \varphi$. Define $(M', \varphi') \geq (M'', \varphi'')$ if and only if $M' \supset M''$ and $\varphi'|_{M''} = \varphi''$. If $(M_i, \varphi_i)_{i \in I}$ is a totally ordered collection of such pairs, then we obtain a map $\bigcup_{i \in I} M_i \to J$ defined by $a \in M_i$ maps to $\varphi_i(a)$. Thus Zorn's lemma applies. To conclude we have to show that if the pair $(M', \varphi')$ is maximal then $M' = N$. In other words, it suffices to show, given any subgroup $M \subset N$, $M \not = N$ and any $\varphi : M \to J$, then we can find $\varphi' : M' \to J$ with $M \subset M' \subset N$ such that (a) the inclusion $M \subset M'$ is strict, and (b) the morphism $\varphi'$ extends $\varphi$.

To prove this, pick $x \in N$, $x \not \in M$. Let $I = \{f \in R \mid fx \in M\}$. This is an ideal of $R$. Define a homomorphism $\psi : I \to J$ by $f \mapsto \varphi(fx)$. Extend to a map $\tilde\psi : R \to J$ which is possible by assumption (3). By our choice of $I$ the kernel of $M \oplus R \to J$, $(y, f) \mapsto y - \tilde\psi(f)$ contains the kernel of the map $M \oplus R \to N$, $(y, f) \mapsto y + fx$. Hence this homomorphism factors through the image $M' = M + Rx$ and this extends the given homomorphism as desired. $\square$

The code snippet corresponding to this tag is a part of the file `more-algebra.tex` and is located in lines 12115–12125 (see updates for more information).

```
\begin{lemma}
\label{lemma-characterize-injective-bis}
Let $R$ be a ring. Let $J$ be an $R$-module.
The following are equivalent
\begin{enumerate}
\item $J$ is injective,
\item $\Ext^1_R(R/I, J) = 0$ for every ideal $I \subset R$, and
\item for an ideal $I \subset R$ and module map $I \to J$
there exists an extension $R \to J$.
\end{enumerate}
\end{lemma}
\begin{proof}
We have seen the implication (1) $\Leftrightarrow$ (2) in
Lemma \ref{lemma-characterize-injective}. In this proof
we will show that (1) $\Leftrightarrow$ (3) which is known
as Baer's criterion.
\medskip\noindent
Assume (1). Given a module map $I \to J$ as in (3) we find
the extension $R \to J$ because the map
$\Hom_R(R, J) \to \Hom_R(I, J)$ is surjective
by definition.
\medskip\noindent
Assume (3). Let $M \subset N$ be an inclusion of $R$-modules.
Let $\varphi : M \to J$ be a homomorphism. We will show that $\varphi$
extends to $N$ which finishes the proof of the lemma.
Consider the set of homomorphisms $\varphi' : M' \to J$
with $M \subset M' \subset N$ and $\varphi'|_M = \varphi$.
Define $(M', \varphi') \geq (M'', \varphi'')$ if
and only if $M' \supset M''$ and $\varphi'|_{M''} = \varphi''$.
If $(M_i, \varphi_i)_{i \in I}$ is a totally
ordered collection of such pairs, then we obtain a map
$\bigcup_{i \in I} M_i \to J$ defined by $a \in M_i$
maps to $\varphi_i(a)$. Thus Zorn's lemma applies.
To conclude we have to show that if the pair
$(M', \varphi')$ is maximal then $M' = N$.
In other words, it suffices to show, given
any subgroup $M \subset N$, $M \not = N$ and
any $\varphi : M \to J$, then we can find
$\varphi' : M' \to J$ with $M \subset M' \subset N$
such that (a) the inclusion $M \subset M'$ is strict, and
(b) the morphism $\varphi'$ extends $\varphi$.
\medskip\noindent
To prove this, pick $x \in N$, $x \not \in M$.
Let $I = \{f \in R \mid fx \in M\}$. This is an ideal of $R$.
Define a homomorphism $\psi : I \to J$ by $f \mapsto \varphi(fx)$.
Extend to a map $\tilde\psi : R \to J$ which is possible by assumption (3).
By our choice of $I$ the kernel of
$M \oplus R \to J$, $(y, f) \mapsto y - \tilde\psi(f)$
contains the kernel of the map $M \oplus R \to N$,
$(y, f) \mapsto y + fx$. Hence this homomorphism factors
through the image $M' = M + Rx$ and this extends the given homomorphism
as desired.
\end{proof}
```

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